My Math Forum how can these to be factor?

 Calculus Calculus Math Forum

 September 13th, 2009, 04:54 PM #1 Newbie   Joined: Sep 2009 Posts: 1 Thanks: 0 how can these to be factor? 3x^2 - 7x + 3 and 2x^3 + 128 i have no idea how the first one can be factor... i dont think it can be factor... the second one, i went: 2[(x^3 + 64)] could this be answer? seems like i could go more steps
 September 14th, 2009, 05:02 AM #2 Senior Member   Joined: Mar 2009 Posts: 318 Thanks: 0 To learn how to factor quadratics (something they were supposed to have covered back in early algebra), try here. For the second one, you need now to apply the difference-of-cubes factoring formula. (Memorize this, along with the sum-of-cubes formula and the difference-of-squares formula. You will need all of these!)
 September 14th, 2009, 09:25 PM #3 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 We can possibly factor 3x² - 7x + 3 using irrational numbers. Let the given expression equal to 0, so 3x² - 7x + 3 = 0. Using quadratic formula for this equation, we arrive with roots of x = (7 ± ?(13)) / 6. Since 3x² - 7x + 3 = 0 is equivalent to x² - (7/3)x + 1 = 0, it's clear to see that (x - (7 + ?(13)) / 6)(x - (7 - ?(13)) / 6) = 0, then it's obvious that 3(x - (7 + ?(13)) / 6)(x - (7 - ?(13)) / 6) = 3x² - 7x + 3 = 0. ? 3(x - (7 + ?(13)) / 6)(x - (7 - ?(13)) / 6). To factor the expression 2x³ + 128 (or 2(x³ + 4³)), use the suggested link given by stapel, we have 2(x + 4)(x² - 4x + 16).

 Tags factor

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Albert.Teng Algebra 7 June 16th, 2012 04:30 AM daigo Algebra 3 June 14th, 2012 05:38 PM HellBunny Algebra 3 February 18th, 2012 10:31 AM Eminem_Recovery Algebra 11 June 19th, 2011 08:50 PM profetas Algebra 3 August 25th, 2009 09:20 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top